Symmetry of Laplace operator

Question

Good evening, I have a question regarding the proof of the symmetry property of Laplace operator. It's a well known fact that $-\Delta$ is an unbounded operator on $L^2(\mathbb{R}^n)$ and it is densely defined with domain $D(-\Delta)=\{f\in L^2:-\Delta f\in L^2\}$. I want to prove that $-\Delta$ is symmetric. It is clear that $\forall f\in\mathcal{S}'$ and $\forall g\in \mathcal{S}$ (here $\mathcal{S}$ denotes the space of Schwartz functions), \begin{equation} \langle-\Delta f,g\rangle=\langle f,-\Delta g\rangle \end{equation} where $\langle\cdot,\cdot\rangle$ denotes the duality pairing $\mathcal{S}'$ / $\mathcal{S}$. Thus, the identity holds whenever $f\in D(-\Delta)$ and $\forall g\in\mathcal{S}$, since $D(-\Delta)\subseteq\mathcal{S}'$. In this case, \begin{equation} (-\Delta f,g)_2=(f,-\Delta g)_2 \ \ \ \forall f\in D(-\Delta),g\in\mathcal{S}. \end{equation} The question is: how do I extend the relation to $g\in D(-\Delta)$? Every text I find claims that it comes from the density of $\mathcal{S}$ in $D(-\Delta)$, but $-\Delta$ in unbounded, so how do I use the density argument here if I can't pass the limit through $-\Delta$? I mean, if $g\in D(-\Delta)$, then I can take a sequence $(g_k)_k\subset\mathcal{S}$ s.t. $\Vert g-g_k\Vert_{L^2}\rightarrow0$. However, \begin{equation} (-\Delta f,g)_{L^2}=\lim_{k}(-\Delta f,g_k)_{L^2}=\lim_k(f,-\Delta g_k)_{L^2}=(f,-\lim_k\Delta g_k)_{L^2}\underset{why?}{=}(f,-\Delta \lim_k g_k)_{L^2}. \end{equation}